Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 26}{x - 5} = \dfrac{11x - 4}{x - 5}$
Solution: Multiply both sides by $x - 5$ $ \dfrac{x^2 + 26}{x - 5} (x - 5) = \dfrac{11x - 4}{x - 5} (x - 5)$ $ x^2 + 26 = 11x - 4$ Subtract $11x - 4$ from both sides: $ x^2 + 26 - (11x - 4) = 11x - 4 - (11x - 4)$ $ x^2 + 26 - 11x + 4 = 0$ $ x^2 + 30 - 11x = 0$ Factor the expression: $ (x - 6)(x - 5) = 0$ Therefore $x = 6$ or $x = 5$ At $x = 5$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 5$, it is an extraneous solution.